Cho các số nguyên $\it{a},\,\it{b},\,\it{c},\,\it{d}$ sao cho $\it{ad}= \it{b}^{\,\it{2}}+ \it{bc}+ \it{c}^{\,\it{2}}$ .
Chứng minh rằng: $\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}+ \it{d}^{\,\it{2}}$ không là hợp số.
$\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}+ \it{d}^{\,\it{2}}= \left ( \it{a}+ \it{b}+ \it{c}+ \it{d} \right )\left ( \it{a}+ \it{d}- \it{b}- \it{c} \right )+ \it{2}\,\left ( \it{b}^{\,\it{2}}+ \it{bc}+ \it{c}^{\,\it{2}}- \it{ad} \right )=$ $\left ( \it{a}+ \it{b}+ \it{c}+ \it{d} \right )\left ( \it{a}+ \it{d}- \it{b}- \it{c} \right )$
Hiển nhiên: $\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}+ \it{d}^{\,\it{2}}\geqq$ $\max\left \{ \left | \it{a}+ \it{b}+ \it{c}+ \it{d} \right |,\,\left | \it{a}+ \it{d}- \it{b}- \it{c} \right | \right \}$
Bài toán trên chỉ đúng với:
$\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}+ \it{d}^{\,\it{2}}=$ $\max\left \{ \left | \it{a}+ \it{b}+ \it{c}+ \it{d} \right |,\,\left | \it{a}+ \it{d}- \it{b}- \it{c} \right | \right \}$ , khi đó:
$$\it{a},\,\it{b},\,\it{c},\,\it{d}\in \left \{ \it{0},\,\it{1} \right \}$$
$\lceil$ Biệt thức $-\,\it{3}$ $\rfloor$